Question

When 3-methylbutan-2-ol is treated with HBr, the following reaction takes place:
Give a mechanism for this reaction.
(Hint : The secondary carbocation formed in step II rearranges to a more stable tertiary carbocation by a hydride ion shift from 3rd carbon atom.)

Answer 1

The mechanism for the reaction between $3$-methylbutan-$2$-ol and HBr is shown above.
In the first step, $-OH$ group is protonated. In the second step, a molecule of water is lost to form secondary carbocation.
Third step is the rearrangement of less stable secondary carbocation to more stable tertiary carbocation through $1,2$-hydride shift.
Final step is the nucelophilic attack of bromide ion on tertiary carbocation to form $2$-bromo-$2$-methylbutane.